I ma given the following sets $X_a = \{(x,y,z) \in \mathbb{R}^3 | ax = y^2 + z^2\}$ and $X_b = \{(x,y,z) \in \mathbb{R}^3 | y^2 + bz= 1\}$. I am asked for which values of $a$ and $b$ are $X_a$ a smooth 2-manifold in $\mathbb{R}^{3}$, for which values is $Y_b$ a smooth 2-manifold in $\mathbb{R}^3$, for which values of $a$ and $b$ is the intersection $X_a \hat Y_b$ a smooth 1-manifold in $\mathbb{R}^3$?
My attempt at the first two questions is the following: For $X_a$, we can define the function $F(x,y,z) = ax -y^2-z^2$. Basically, we want to find the values of $a$ for which $0$ would be a regular value. THus, we compute the differential of the function $F$, which is gonna be given by the matrix $(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}) = (a, -2y, -2z)$. Now, we know $0$ is gonna be a regular value iff $\forall p \in F^{-1}(0)$, the matrix $(a, -2y, -2z)$ evaluated at this point is not all entries 0, since then for any such value the differential will be surjective. But this will happen only for the point $(0,0,0) \in \mathbb{R}^3$. Thus, we conclude if $a \in \mathbb{R}\setminus \{0\}$ then $X_a$ will be a smooth $2$-manifold in $\mathbb{R}^3$. (Of course this is using Regular value theorem).
For the second question, I did a similar process, but this time the differential of the function $F(x,y,z) = y^2+bz-1$ is given by $(0,2y,b)$. With the exact same reasoning, we conclude that $Y_b$ is gonna be a smooth $2$-manifold in $\mathbb{R}^3$ whenever $b\in \mathbb{R} \setminus \{0\}$.
For the third part, I am trying a similar approach, this time I am defining the map $F(x,y,z) = (ax-y^2- z^2, y^2+bz-1)$. Now, the differential of this map is the matrix
\begin{bmatrix} a & -2y & -2z \\ 0 & 2y & b \end{bmatrix}
Now, again we want to find the values of $a$ and $b$ such that the differential is not surjective. So, we need to find the values of $a$ and $b$ that make the matrix above to have rank less than 2. So we need to find the values of $a$ and $b$ such that all the determinant of all $2 \times 2$ matrices vanish. By calculation, we end up having that these determinants are $2ya=0$, $ab=0$ and $-2yb + 4yz=0$. From this we conclude that the for $a=0, b=2yz$, the matrix will have rank less than 2 and so for $a \in \mathbb{R}\setminus \{0\}$ and $b \in \mathbb{R} \setminus \{b=2yz\}$, the intersection $X_a \cap Y_b$ would be a smooth 1-manifold in $\mathbb{R}^3$
Can anybody tell me if my process is correct or if not, where I could correct the mistakes? Thanks!